Tài liệu Physics exercises_solution: Chapter 13 ppt

13.1: a) s.rad1038.12s,1055.4 3 2 3 1   πfωT T π f b) s.rad1053.52s,1014.1 33 Hz)220(4 1   πfω 13.2: a) Since the glider is released form rest, its initial displacement (0.120 m) is the amplitude. b) The glider will return to its original position after another 0.80 s, so the period is 1.60 s. c) The frequency is the reciprocal of the period (Eq. (13.2)),  s60.1 1 f Hz.625.0 13.3: The period is s1014.1 3 440 s50.0   and the angular frequency is  T π ω 2 s.rad1053.5 3  13.4: (a) From the graph of its motion, the object completes one full cycle in 2.0 s; its period is thus 2.0 s and its frequency .s5.0period1 1  (b) The displacement varies from m,20.0 tom20.0  so the amplitude is 0.20 m. (c) 2.0 s (see part a) 13.5: This displacement is 4 1 of a period. s.0500.0sos,200.01  tfT 13.6: The period will be twice the time given as being between the times at which the glider is at the equilibrium position (see Fig. (13.8)); m.N292.0kg)200.0( s)60.2(2 22 2 2 2                  π m T π mωk 13.7: a) kg.084.0c) s.rad7.372b) s.167.0 2 1  ω k f mπfωT 13.8: Solving Eq. (13.12) for k, m.N1005.1 s150.0 2 kg)600.0( 2 3 22                π T π mk 13.9: From Eq. (13.12) and Eq. (13.10), Hz,66.2s,375.02 1 mN140 kg500.0  T fπT s.rad7.162  πfω 13.10: a) )(so,)sin( 22 2 2 txxωβωtAωa dt xd x  is a solution to Eq. (13.4) if ωAaω m k 2b). 2  a constant, so Eq. (13.4) is not satisfied. c) , )( βωti dt dx x iωv     mkωtxxωAeiωa βωti dt dv x x 22)(2 if(13.4)Eq.osolution tais)(so,)( 13.11: a) s,m8.29Hz))(440m)(210(3.0b) )Hz))(440((2cosmm)0.3( 3   πtπx ),Hz))(440sin((2)sm1034.6()(c) .sm1029.2Hz)440()mm)(20.3( 372422 tπtjπ  .sm1034.6 37 max j 13.12: a) From Eq. (13.19), m.98.0 00  mk v ω v A b) Equation (13.18) is indeterminant, but from Eq. (13.14), , 2    and from Eq. (13.17), sin .so,0 2 π   c) )).)srad sin((12.2m)98.0(so,sin ))2((cos txωtπωt  13.13: With the same value for ω , Eq. (13.19) gives m383.0A and Eq. (13.18) gives   .rad02.1rad/s)(12.2cosm)(0.383and  tx ,58.5rad02.1 kgN/m/2.00300m)(0.200 m/s)4.00( arctan             and x = (0.383 m) cos ((12.2 rad/s)t + 1.02 rad). 13.14: For SHM,   .m/s71.2m)101.1(Hz)5.2(2)2( 22 2 22   πxπfxωa x b) From Eq. (13.19) the amplitude is 1.46 cm, and from Eq. (13.18) the phase angle is 0.715 rad. The angular frequency is rad/s,7.152 πf so .rad)715.0rad/s)((15.7cos)cm/s359( rad)715.0rad/s)((15.7sin )scm9.22( rad)715.0rad/s)((15.7coscm)46.1( 2    ta tv tx x x 13.15: The equation describing the motion is ;sinωtAx  this is best found from either inspection or from Eq. (13.14) (Eq. (13.18) involves an infinite argument of the arctangent). Even so, x is determined only up to the sign, but that does not affect the result of this exercise. The distance from the equilibrium position is        m.353.054sinm600.02sin  πTtπA 13.16: Empty chair: k m πT 2 N/m993 s)(1.30 kg)5.42(44 2 2 2 2  π T m π k With person in chair: kg120kg5.42kg162 kg162 4 N/m)993(s)54.2( 4 2 person 2 2 2 2    m ππ kT m km πT 13.17: kg400.0,2  mkmπT s09.22 N/m60.3 m300.0 )m/s70.2kg)(400.0( gives :calculate tom/s70.2Use 2 2      kmπT x ma kmakx ka x x x 13.18: We have ).2)s4.71cm/s)sin((60.3()( 1 πttv x   Comparing this to the general form of the velocity for SHM: 2 s4.71 cm/s60.3 1 π ω ωA     (a) s33.1s71.422T 1   πωπ (b) cm764.0 s71.4 scm60.3scm60.3 1   ω A (c ) 2212 max scm9.16)cm764.0()s71.4(   Aωa 13.19: rad)42.2s)radcos((4.16cm)40.7()(a)  ttx 22 2222 max max 2 2 1 2 2 1 2 2 1 2 sm216.0sm)0125.0)(50.10.26( .sm0.303isSpeed sm303.0sm)0125.0()0740.0(50.10.26 m0125.0givess00.1at evaluated)(e) N92.1sod) sm308.0gives m0740.0cm40.7c) mN0.26)2(so2b) s1.51so 2s)rad(4.16,When         mkxa xAmkv xttx kAFkxF mkAvkAkxmv A T πmkkmπT T πTTt 13.20: See Exercise 13.15; s.0.0871))(20.36))(1.5arccos((  πt 13.21: a) Dividing Eq. (13.17) by ω , .sin ,cos 0 0  A ω v Ax Squaring and adding, , 2 2 2 0 2 0 A ω v x  which is the same as Eq. (13.19). b) At time ,0t Eq. (13.21) becomes , 2 1 2 1 2 1 2 1 2 1 2 0 2 0 2 2 0 2 0 2 kxv ω k kxmvkA  where 2 kωm  (Eq. (13.10)) has been used. Dividing by 2k gives Eq. (13.19). 13.22: a) s.m1.48m)10Hz))(0.60392(2()2( 3 max   πAπfv b) J.1096.2s)mkg)(1.48107.2( 2 1 )( 2 1 5252 maxmax   VmK 13.23: a) Setting 2 2 1 2 2 1 kxmv  in Eq. (13.21) and solving for x gives . 2 A x  Eliminating x in favor of v with the same relation gives .2 2 2 ωA x mkAv  b) This happens four times each cycle, corresponding the four possible combinations of + and – in the results of part (a). The time between the occurrences is one-fourth of a period or   8 3 84 3 4 1 24 2 22 , ,c) .4/ kAkA ω π ω π KUEKEUT  13.24: a) From Eq. (13.23), m/s.1.20m)040.0( kg0.500 mN450 max  A m k v b) From Eq. (13.22), m/s.11.1m)015.0(m)040.0( kg0.500 N450 22 v c) The extremes of acceleration occur at the extremes of motion, when ,Ax  and 2 max m/s36 kg)(0.500 m)N/m)(0.040450(  m kA a d) From Eq. (13.4), .m/s5.13 2 kg)(0.500 m)0.015N/m)(450(   x a e) From Eq. (13.31), J. 36.0m)N/m)(0.040450( 2 2 1 E 13.25: a)     max 22 2 22 max .m/s5.13m)100.18(Hz)85.0(2)2( vAfAωa  m/s961.02  πfAωA . ,m/s57.2)2(b) 22  xπfa x   m/s.833.0m)100.9(m)100.18(Hz)85.0(2 )2( 2222 22    π xAπfv c) The fraction of one period is )21( π arcsin ),0.180.12( and so the time is )2( πT arcsin 1 1037.1)0.180.12(   s. Note that this is also arcsin ωAx )( . d) The conservation of energy equation can be written 2 2 1 2 2 1 2 2 1 kxmvkA  . We are given amplitude, frequency in Hz, and various values of x . We could calculate velocity from this information if we use the relationship 222 4 fπωmk  and rewrite the conservation equation as 2 2 1 4 2 1 2 2 1 22 2 xA fπ v  . Using energy principles is generally a good approach when we are dealing with velocities and positions as opposed to accelerations and time when using dynamics is often easier. 13.26: In the example, mM M AA   12 and now we want ,So. 2 1 1 2 1 2 mM M AA   or Mm 3 . For the energy, 2 2 2 1 2 kAE  , but since 1 4 3 1 4 1 21 2 1 2 or ,, EEEAA  is lost to heat. 13.27: a) J 0284.0 2 2 1 2 2 1  kxmv . b) m.014.0 kg)(0.150N/m)300( m/s)300.0( m)012.0( 2 2 2 2 0 2 0  ω v x c)  sm615.0mAkωA 13.28: At the time in question we have 22 sm40.8)(cos sm20.2)sin( m600.0)(cos       ωtAωa ωtωAv ωtAx Using the displacement and acceleration equations: 222 sm40.8m)600.0()(cos  ωωtAω  12 s742.3and0.14   ωω To find A, multiply the velocity equation by :ω 212 sm232.8)sm(2.20)s742.3()(sin    ωtAω Next square both this new equation and the acceleration equation and add them: m840.0 m7054.0 )s742.3( sm3.138sm3.138 sm3.138sm56.70sm77.67 )(cos)(sin )sm40.8()sm232.8()(cos)(sin 2 41 42 4 42 2 42424224 2224 2222224224       A ω A A ω ωtωt Aω ωt AωωtAω   The object will therefore travel m0.240m600.0m840.0  to the right before stopping at its maximum amplitude. 13.29: mkAv  max sm509.0Then m.0405.0)(so :find to Use s158)2(so2 :find to Use max maxmax max 22     mkAv mkaAmkAa Aa TπmkkmπT mkT 13.30: Using 0 0 L F k  from the calibration data, kg.00.6 Hz))60.2((2 m)10(1.25N)200( )2( )( 2 1 2 00     ππf LF m 13.31: a) m.N10153 m)120.0( )sm(9.80kg)650( Δ 3 2  l mg k b) s.695.0 sm9.80 m120.0 222 2  π g l π k m πT  13.32: a) At the top of the motion, the spring is unstretched and so has no potential energy, the cat is not moving and so has no kinetic energy, and the gravitational potential energy relative to the bottom is J 3.92m)050.0()m/skg)(9.8000.4(22 2 mgA . This is the total energy, and is the same total for each part. b) J 92.3so,0,0 springgrav  UKU . c) At equilibrium the spring is stretched half as much as it was for part (a), and so J 98.0soandJ,1.96J)92.3(J, 0.98J)92.3( 2 1 grav 4 1 spring  KUU . 13.33: The elongation is the weight divided by the spring constant, cm97.3 4 2 2 2  π gT mω mg k w l  . 13.34: See Exercise 9.40. a) The mass would decrease by a factor of 271)31( 3  and so the moment of inertia would decrease by a factor of )2431()31)(271( 2  , and for the same spring constant, the frequency and angular frequency would increase by a factor of 6.15243  . b) The torsion constant would need to be decreased by a factor of 243, or changed by a factor of 0.00412 (approximately). 13.35: a) With the approximations given, ,mkg1072.2 282   mRI 28 mkg102.7or   to two figures. b) radmN103.4)mkg1072.2(Hz)22()2( 62822   πIπfκ . 13.36: Solving Eq. (13.24) for  in terms of the period, m/rad.N1091.1 )m)10kg)(2.201000.2)(21(( s00.1 2 2 5 223 2 2                   π I T π 13.37:   .mkg0152.0 s)(265125)(2 m/radN450.0 )2( 2 2 2      π πf I 13.38: The equation )t(cos φωθ  describes angular SHM. In this problem, .0φ a) ).cos(and)sin( 2 2 2 tωωtωω dt θd dt d θ  b) When the angular displacement is )cos(, tω , and this occurs at ,0t so 1.cos(0)since,and0,sin(0) since 0 2 2 2   ω dt θd dt dθ When the angular displacement is ).cos(or ),cos(,2 2 1 2 tωtω   .21)cos(since, 2 and, 2 3 )sin(since 2 3 2 2 2      tω ω dt θd t ω ω dt dθ This corresponds to a displacement of 60 . 13.39: Using the same procedure used to obtain Eq. (13.29), the potential may be expressed as ].)1(2)1[( 6 0 12 00   RxRxUU Note that at ., 00 UURr  Using the appropriate forms of the binomial theorem for || 0 Rx << 1,                                            2 00 2 00 0 2 76 612 2 1312 121 RxRx RxRx UU        2 2 0 0 36 1 x R U . 2 1 0 2 Ukx  where 2 0 /72 RUk  has been used. Note that terms in 2 u from Eq. (13.28) must be kept ; the fact that the first-order terms vanish is another indication that 0 R is an extreme (in this case a minimum) of U. 13.40:     Hz.1033.1 kg)1066.1(008.1 N/m)580(2 2 1 22 1 14 27       m k f 13.41: ,2 gLT   so for a different acceleration due to gravity ,g    s.60.2sm71.3sm80.9s60.1 22     ggTT 13.42: a) To the given precision, the small-angle approximation is valid. The highest speed is at the bottom of the arc, which occurs after a quarter period, s.25.0 24  g LT  b) The same as calculated in (a), 0.25 s. The period is independent of amplitude. 13.43: Besides approximating the pendulum motion as SHM, assume that the angle is sufficiently small that the length of the spring does not change while swinging in the arc. Denote the angular frequency of the vertical motion as L gkg m k      and 0 , 4 0 2 1 w kg ω  which is solved for kwL 4 . But L is the length of the stretched spring; the unstretched length is     m.00.2N/m50.1N00.133 0  kwkwLL 13.44: 13.45: The period of the pendulum is   s.36.1100s136 T Then,     .sm67.10 s1.36 m5.44 2 2 2 2 2  π T Lπ g 13.46: From the parallel axis theorem, the moment of inertia of the hoop about the nail is   .13.39Eq.in with ,22so,2 222 RdgRπTMRMRMRI  Solving for R, m.496.08 2 2  πgTR 13.47: For the situation described, LdmLI  and 2 in Eq. (13.39); canceling the factor of m and one factor of L in the square root gives Eq. (13.34). 13.48: a) Solving Eq. (13.39) for I,       .mkg0987.0m250.0sm80.9kg1.80 2 s940.0 2 22 22                π mgd π T I b) The small-angle approximation will not give three-figure accuracy for rad.0.400Θ  From energy considerations,   .Ω 2 1 cos1 2 max Imgd  Expressing max  in terms of the period of small-angle oscillations, this becomes      .srad66.2rad0.40cos1 s940.0 2 2cos1 2 2 22 max                π T π 13.49: Using the given expression for I in Eq. (13.39), with d=R (and of course m=M), s.58.0352  gRπT 13.50: From Eq. (13.39),       .kg.m129.0 2 100s120 m200.0sm9.80kg80.1 2 2 2 2 2                ππ T mgdI 13.51: a) From Eq. (13.43),         Hz.393.0 2 so,srad47.2 kg300.04 skg90.0 kg300.0 mN50.2 2 2       π ω fω b)    .skg73.1kg300.0mN50.222  kmb 13.52: From Eq. (13.42)   ,for Solving.exp 2 12 btAA m b  s.kg0220.0 m100.0 m300.0 ln )s00.5( )kg050.0(2 ln 2 2 1                    A A t m b As a check, note that the oscillation frequency is the same as the undamped frequency to valid.is(13.42)Eq.so,%108.4 3  [...]... b  2km and positive if b  2km The graph in the three cases will be curved down, not curved, or curved up, respectively 13. 54: At resonance, Eq (13. 46) reduces to A  Fmax bd a) 31 b) 2 A1 Note that the resonance frequency is independent of the value of b (see Fig (13. 27)) A 13. 55: a) The damping constant has the same units as force divided by speed, or kg  m s 2 m s   kg s  b)The units... c) ωd  k m (i) bωd  0.2 k , so A  Fmax 0.2k   5 Fmax k (ii) bd  0.4k , so A  Fmax (0.4 k )  2.5Fmax k , as shown in Fig. (13. 27) 13. 56: The resonant frequency is k m  (2.1  106 N m) 108 kg )  139 rad s  22.2 Hz, and this package does not meet the criterion 13. 57: a) 2  π rad s    0.100 m   2 3 a  Aω      6.72  10 m s   (3500 rev min )   2  30 rev min    2  π rad... half the amplitude is sin ωt  1 2, or ωt  π 6 rad, or one-twelfth of a period The needed time is twice this, or one-sixth of a period, 0.70 s d) l  mg  ωg2  2πgr 2  4.38 m k 13. 69: 2 13. 70: Expressing Eq (13. 13) in terms of the frequency, and with   0, and taking two derivatives,  2πt  x  0.240 m  cos   1.50 s      2π 0.240 m    2πt   2πt  v x    1.50 s   sin ... )  44.5 N 13. 75: Of the many ways to find the time interval, a convenient method is to take 1   0 in Eq (13. 13) and find that for x  A 2, cos ω t  cos(2πt / T )  2 and so t  T / 6 The time interval available is from  t to t , and T / 3  1.17 s See Problem 12.84; using x as the variable instead of r , dU GM E GM E m g F ( x)    x, so ω2   3 3 dx RE RE RE The period is then 13. 76: T... 1.31 c  1  u 4  A  32m T , A0 which may be expressed as T  motion is not simple harmonic 7.41 A m c c) The period does depend on amplitude, and the 13. 79: As shown in Fig 13. 5b , v  vtan sinθ With vtan  Aω and θ  ωt  , this is Eq 13. 15 13. 80: a) Taking positive displacements and forces to be upwad, 2 n  mg  ma, a  2πf  x, so   n  m g  2πf  A cos2πf  t    2 a) The fact... comparison to the length of the rod; see Problem 13. 94) k 13. 92: Using the notation 2bm  γ, m  ω2 and taking derivatives of Eq (13. 42) (setting the phase angle   0 does not affect the result), x  Αe  t cosω t vx   Αe  t (ω sin ω t   cos ω t ) ax   Αe  t ((ω 2  γ 2 ) cos ω t  2ω  sin ω t ) Using these expression in the left side of Eq (13. 41),  kx  bvx  Αe  t (  k cos ω... the beam is 0.426 m below the equilibrium position   13. 73: The pendulum swings through 1 cycle in 1.42 s, so T  2.84 s L  1.85 m 2 Use T to find g: 2 T  2π L g so g  L2π T   9.055 m/s 2 Use g to find the mass M p of Newtonia: 2 g  GM p / Rp 2πRp  5.14  107 m, so Rp  8.18  106 m 2 gRp mp   9.08  1024 kg G 13. 74: a) Solving Eq (13. 12) for m , and using k  2 F l 2  T  F  1  40.0... 13. 10 13. 88: The torque on the rod about the pivot (with angles positive in the direction indicated in the figure) is τ  k L θ  L Setting this equal to the rate of change of angular 2 2 momentum, I  I d 2 dt 2 , d 2θ L2 4 3k  k θ   θ, 2 dt I M 1 where the moment of inertia for a slender rod about its center, I  12 ML2 has been used It follows that ω2  3K M , and T  2π ω  2π M 3k 13. 89:... be the same as that of the bell; equating the expression in Eq (13. 34) to that in Eq (13. 39) and solving for L gives L  Ι md  (18.0 kg  m 2 ) ((34.0 kg)(0.60 m))  0.882 m Note that the mass of the bell, not the clapper, is used As with any simple pendulum, the period of small oscillations of the clapper is independent of its mass 13. 90: The moment of inertia about the pivot is 2(1 3) ML2  (2 3)... distance d  L (2 2 ) below the pivot (see Problem 8.95) From Eq (13. 39), the frequency is f  1 1  T 2π 3g 1  4 2 L 4π 3g  2L 13. 91: a) L  g (T 2π ) 2  3.97 m b)There are many possibilities One is to have a uniform thin rod pivoted about an axis perpendicular to the rod a distance d from its center Using the desired period in Eq (13. 39) gives a quadratic in d, and using the maximum size for the . mπfωT 13. 8: Solving Eq. (13. 12) for k, m.N1005.1 s150.0 2 kg)600.0( 2 3 22                π T π mk 13. 9: From Eq. (13. 12) and Eq. (13. 10),. max j 13. 12: a) From Eq. (13. 19), m.98.0 00  mk v ω v A b) Equation (13. 18) is indeterminant, but from Eq. (13. 14), , 2    and from Eq. (13. 17),